Bài 2: 

\(=\dfrac{28}{25}\cdot\dfrac{15}{7}\cdot5=\dfrac{75}{25}\cdot4=12\)

Bài 1: 

a: \(x+\dfrac{7}{8}=\dfrac{13}{2}:4=\dfrac{13}{8}\)

nên x=13/8-7/8=6/8=3/4

b: \(x:\dfrac{5}{3}=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18-10}{15}=\dfrac{8}{15}\)

nên \(x=\dfrac{8}{15}\cdot\dfrac{5}{3}=\dfrac{8}{9}\)

Bài 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

\(x=\dfrac{1}{3}\)

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

\(4-\left(2x+1\right)=\dfrac{8}{3}\)

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

\(2x+1=\dfrac{20}{3}\)

\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

\(\left(x+4\right)^3=\left(-5\right)^3\)

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

Uh, chừa sau k dám học muộn nx

Đăng từng bài `1` thôi cậu ơii

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

Câu hỏi của chíp chíp - Toán lớp 7 | Học trực tuyến Full dễ ok??

OK???????

Trừ câu F ra thì dệ hết

ai làm dùm tôi thì làm đi. Ngồi đó kêu dễ.

Mk ms làm xong có câu D ak Mashiro Shiina

Làm mk 2 câu còn lại đi Mashiro Shiina

1

a)103/35

b)-3/13

\(a,\dfrac{-8}{15}+\dfrac{13}{30}-\dfrac{5}{12}=\dfrac{-32}{60}+\dfrac{26}{60}-\dfrac{25}{60}=-\dfrac{31}{60}\\ b,\dfrac{3}{2}.\dfrac{7}{2}+\left(\dfrac{-5}{6}+\dfrac{1}{10}:\dfrac{11}{30}\right)=\dfrac{21}{4}+\left(\dfrac{-5}{6}+\dfrac{3}{11}\right)=\dfrac{21}{4}+\dfrac{-37}{66}=\dfrac{619}{132}\)

\(c,\dfrac{-20}{21}.\dfrac{22}{35}+\dfrac{-20}{21}.\dfrac{13}{35}+\dfrac{-22}{21}=\dfrac{-20}{21}\left(\dfrac{22}{35}+\dfrac{13}{35}\right)+\dfrac{-22}{21}=\dfrac{-20}{21}.1+\dfrac{-22}{21}=\dfrac{-20}{21}+\dfrac{-22}{21}=\dfrac{-42}{21}=-2\)

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)

\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)

Bài 2: 

a: \(=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}=\dfrac{-6}{9}=\dfrac{-2}{3}\)

b: \(=\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{10}{40}-\dfrac{10}{56}+\dfrac{10}{104}}\)

\(=\dfrac{\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{\dfrac{10}{8}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{2}{3}:\dfrac{5}{4}=\dfrac{2}{3}\cdot\dfrac{4}{5}=\dfrac{8}{15}\)

c: \(=\dfrac{2\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}{4\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{13\left(\dfrac{5}{121}-\dfrac{2}{21}-1\right)}\)

=2/4:(-1)/13=2/4x(-13)=-13/2

bài 1

a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)

d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)

bài 2

a)\(x=\dfrac{5}{6}-\dfrac{4}{5}=\dfrac{25}{30}-\dfrac{24}{30}=\dfrac{1}{30}\)

b)\(x=5\times\dfrac{10}{7}=\dfrac{50}{7}\)

bài 4 :

145  69 + 22 x  145 +145 x 8 + 145 

\(=145\times\left(69+22+8+1\right)=145\times100=14500\)

bài 1:

a, \(\dfrac{2}{5}+\dfrac{3}{8}=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b,\(\dfrac{7}{6}-\dfrac{2}{3}=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c,\(\dfrac{5}{9}x6=\dfrac{5}{9}x\dfrac{6}{1}=\dfrac{30}{9}\)

d,\(\dfrac{8}{5}:\dfrac{4}{7}=\dfrac{8}{5}x\dfrac{7}{4}=\dfrac{14}{5}\)

bài 2 :

\(a,\dfrac{4}{5}+x=\dfrac{5}{6}\)

            \(x=\dfrac{5}{6}-\dfrac{4}{5}\)

            \(x=\dfrac{1}{30}\)

b, \(x:\dfrac{7}{10}=5\)

    \(x\)         \(=5x\dfrac{7}{10}\)

    \(x\)         \(=\dfrac{35}{10}\)    

bài 3 :

đổi :16 tấn 8 tạ = 168 tạ

        2 tấn 6 tạ = 26 tạ 

xe ô tô thứ nhất chở số tạ hàng là:

      ( 168 + 26 ) : 2= 97 ( tạ)

xe ô tô thứ hai chở số tạ hàng là:

         97 - 26 = 71 ( tạ)

               đáp số :xe ô tô thứ nhất : 97 tạ thóc

                            xe ô tô thứ hai  : 71 tạ thóc 

ai giúp mik ik T_T

a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))

= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))

= \(\dfrac{1}{13}\) \(\times\)  \(\dfrac{317}{73}\)

= \(\dfrac{317}{949}\)

b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)

=   \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)

= \(\dfrac{199}{45}\)

c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

=   \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)

= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)

= \(\dfrac{20696}{315}\)

d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)

= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)

= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))

= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)

= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)

e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))

= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)

=  -1 - \(\dfrac{2}{11}\)  - \(\dfrac{8}{33}\)

= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) -  \(\dfrac{8}{33}\)

= - \(\dfrac{47}{33}\)